Answer:
second choice
Step-by-step explanation:
The given sums ∑ have j = 1, j = 2, j = 3, j = 4, j = 5
Consider first j = 1 and substitute in the answers to see if we get the first term of the series -3
[tex]-3(\frac{1}{3} )^{1-1} = -3(\frac{1}{3} )^{0}= -3*1= -3[/tex] ✅
2·1 - 5 = 2-5 = -3 ✅
3·1 - 6 = 3-6 = -3 ✅
[tex]5^{1-1} = 5^{0} = 1[/tex] ❌
We see that last choice gives us as first term to be 1 not -3 so eliminate that possibility.
Consider now j = 2 and substitute in the answers to see if we get the second term of the series -1
[tex]-3(\frac{1}{3} )^{2-1} = -3(\frac{1}{3} )^{1}= \frac{-3}{3} = -1[/tex] ✅
2·2 - 5 = 4-5 = -1 ✅
3·2 - 6 = 6-6 = 0 ❌
We see that third choice gives us as second term to be 0 not -1 so eliminate that possibility.
Consider now j = 3 and substitute in the answers to see if we get the third term of the series 1
[tex]-3(\frac{1}{3} )^{3-1} = -3(\frac{1}{3} )^{2}= \frac{-3}{9} = \frac{-1}{3}[/tex] ❌
2·3 - 5 = 6-5 = 1 ✅
We see that first choice gives us as third term to be -1/3 not 1 so eliminate that possibility.
Now we are left with the second choice as the only possibility.
